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Find it's value


Sriman Dutta

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Irrelevant link.

 

[math]a^a=b \Rightarrow a=\log_a b =(^a\sqrt{b})[/math].

The link is fully relevant, he's trying to find "a" as a function of "b". You can't do that.

 

 

 

Of course this generalizes - [math]x^y=z \Rightarrow y=log_xz\,\,\ne \,\,(^y\sqrt{z})[/math]

 

This is a DIFFERENT problem. Do you see the difference?

Edited by zztop
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